How do you find the axis of symmetry, vertex and x intercepts for y=-x^2-3x+4?

Jul 18, 2017

axis of symmetry is $x = - \frac{3}{2}$
vertex minimum at $\frac{25}{4}$
$x$-intercepts are $\left(1 , 0\right) \mathmr{and} \left(- 4 , 0\right)$

Explanation:

$y = - {x}^{2} - 3 x + 4$

to find axis of symmetry and vertex, we can use completing a square to solve it.

$y = - \left({x}^{2} + 3 x\right) + 4$

$y = - {\left(x + \frac{3}{2}\right)}^{2} + {\left(\frac{3}{2}\right)}^{2} + 4$

$y = - {\left(x + \frac{3}{2}\right)}^{2} + \frac{9}{4} + \frac{16}{4}$

$y = - {\left(x + \frac{3}{2}\right)}^{2} + \frac{25}{4}$ $\to i$

it axis of symmetry can be found when $\left(x + \frac{3}{2}\right) = 0$, then $x = - \frac{3}{2}$

since $- {\left(x + \frac{3}{2}\right)}^{2}$ is always has a -ve value, therefore it has a maximum vertex at $\frac{25}{4}$

to find $x$-intercepts, plug in $y = 0$ in $\to i$

$0 = - {\left(x + \frac{3}{2}\right)}^{2} + \frac{25}{4}$

${\left(x + \frac{3}{2}\right)}^{2} = \frac{25}{4}$

$\left(x + \frac{3}{2}\right) = \pm \sqrt{\frac{25}{4}}$

$x = - \frac{3}{2} \pm \frac{5}{2}$

${x}_{1} = - \frac{3}{2} + \frac{5}{2} = \frac{2}{2} = 1$

${x}_{2} = - \frac{3}{2} - \frac{5}{2} = - \frac{8}{2} = - 4$

therefore $x$-intercepts are $\left(1 , 0\right) \mathmr{and} \left(- 4 , 0\right)$